Prove that $\int_0^t\frac{(B^1_s)^2}{\big((B_s^1)^2+(B_s^2)^2+(B_s^3)^2\big)^3}ds<\infty $ a.s.

Question

Let $(B^1,B^2,B^3)$ a Brownian motion in $\mathbb R^3$. Prove that $$\int_0^t\frac{(B^1_s)^2}{\big((B_s^1)^2+(B_s^2)^2+(B_s^3)^2\big)^3}ds<\infty \quad a.s.$$

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I'm not so sure how to do it. Set $f(x,y,z)=\frac{x^2}{(x^2+y^2+z^2)^3}$. I proved that $$m\{s\in [0,t]\mid (B_s^1)^2+(B_s^2)^2+(B_s^3)^2=0\}=0\ \ a.s.$$ therefore $f(B_s^1,B_s^2,B_s^3)<\infty $ a.s. But unfortunately, this doesn't prove the claim. Moreover, since $f(B_s^1,B_s^2,B_s^3)$ is not a.s. bounded on $[0,t]$, I don't know how to do. Any idea ?

Answer 1

Let $f_1$ be your function. Let $f_2$ and $f_3$ be the same except with $y$ and $z$ on top. By symmetry, it is now enough to find $$\int_0^t \sum_i f_i(B^1_s,B^2_s,B^3_s)ds=\int_0^t ||(B^1_s,B^2_s,B^3_s)||^{-2}ds$$

A nonnegative random variable on a probability space is bounded a.s. iff its expectation is finite. So taking expectation and applying Fubini, we get,

$$\int_0^t\mathbb{E}\left[||(B^1_s,B^2_s,B^3_s)||^{-2}\right]ds$$

So now you just need to understand $$\mathbb{E}\left[||(B^1_s,B^2_s,B^3_s)||^{-2}\right] $$ as a function of s.

Answer 2

If the given integral is finite a.s., then so are the integrals with $(B_s^2)^2$ and $(B_s^3)^2$ in the numerator instead of $(B_1^1)^2$. Adding these three integrals one finds that $$ \int_0^t {1\over \left((B^1_s)^2+(B_s^2)^2+(B_s^3)^2\right)^2}\,ds<\infty, a.s. $$ Now the radial part $R_t:=\sqrt{(B^1_s)^2+(B_s^2)^2+(B_s^3)^2}$ of a 3-dimensional Brownian motion, the so-called 3-dim. Bessel process, has been well studied. In particular, it is known that if $f:(0,\infty)\to(0,\infty)$ is (for simplicity) continuous, then $\int_0^t f(R_s)\,ds<\infty$ for all $t>0$ a.s. if and only if $\int_0^1 rf(r)\,dr<\infty$. [See https://link.springer.com/chapter/10.1007%2FBFb0083762 , "A zero-one law for integral functionals of the Bessel process" by Xing-Xiong Xue.] In the present case, $f(r) = r^{-4}$, for which $\int_0^1 rf(r)\,dr =\int_0^1 r^{-3}\,dr =\infty$. So the Brownian integral in question cannot converge.